2012 AMC 12B Problems/Problem 12

Problem

How many sequences of zeros and ones of length 20 have all the zeros consecutive, or all the ones consecutive, or both?

$\textbf{(A)}\ 190\qquad\textbf{(B)}\ 192\qquad\textbf{(C)}\ 211\qquad\textbf{(D)}\ 380\qquad\textbf{(E)}\ 382$

Solutions

Solution 1

There are $\binom{20}{2}$ selections; however, we count these twice, therefore

$2\cdot\binom{20}{2} = 380$ . The wording of the question implies D not E.

MAA decided to accept both D and E, however.

Solution 2

Consider the 20 term sequence of $0$ 's and $1$ 's. Keeping all other terms 1, a sequence of $k>0$ consecutive 0's can be placed in $21-k$ locations. That is, there are 20 strings with 1 zero, 19 strings with 2 consecutive zeros, 18 strings with 3 consecutive zeros, ..., 1 string with 20 consecutive zeros. Hence there are $20+19+\cdots+1=\binom{21}{2}$ strings with consecutive zeros. The same argument shows there are $\binom{21}{2}$ strings with consecutive 1's. This yields $2\binom{21}{2}$ strings in all. However, we have counted twice those strings in which all the 1's and all the 0's are consecutive. These are the cases $01111...$ , $00111...$ , $000111...$ , ..., $000...0001$ (of which there are 19) as well as the cases $10000...$ , $11000...$ , $111000...$ , ..., $111...110$ (of which there are 19 as well). This yields $2\binom{21}{2}-2\cdot19=\boxed{\textbf{(E)}\ 382}$

2012 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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All AMC 12 Problems and Solutions

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2012 AMC 12B Problems/Problem 12

Contents

Problem

Solutions

Solution 1

Solution 2

See Also